Question

Problem 1 (15 points 4 + 4 + 4 + 3] Educators conducted a study of how students' involvement into social networks = X measured in hours per week affects their test scores =Y (in a 100-point scale). They assumed all conditions required for this model and used a sample of n = 20 students. As a result, they derived the fitted line equation in the form: û = 70 – 2.85 · X with the standard error for the estimated slope equal to SE [b] 1.5 at social networking has NEGATIVE At the significance vel a = 0.05, do they have enough evidence impact on students' test performance? 1. Specify the critical value (or values) required in this case. 2. Formulate Rejection Rule 3. Evaluate the Test Statistic 4. Finally, state your decision. Solution

Answer 1

Answer Date: 27/07/2020 To test the hypothesis is that social networking has negative impact on students' test performance at 5% significance level. The null and alternative hypothesis is, Ho:B. = 0 H.B. < 0 1) The t critical value is, The sample size is small and left-tailed test. Look in the column headed a = 0.05 and the row headed in the t distribution table by using degree of freedom is, d.f.=n-2 = 20-2 = 18 The t critical value is - 1.734

2) The rejection rule is, Reject the null hypothesis, if the t-test statistics is less than the t-test statistics which is 1.734. 3) The t-test statistics is, Coefficient -0 t= S.E.Coefficients - 2.85-0 1.50 = -1.900 The t-test statistics is -1.900 4)

Decision The statistical decision is that the t-test statistics which is -1.900 is less than t-critical value which is -1.734, so the null hypothesis is rejected at 5% level of significance. There is sufficient evidence to indicate that social networking has negative impact on students' test performance. The result is statistically significant.